q^2+10q-56=0

Solution for q^2+10q-56=0 equation:

q^2+10q-56=0

a = 1; b = 10; c = -56;
Δ = b2-4ac
Δ = 102-4·1·(-56)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-18}{2*1}=\frac{-28}{2}
=-14
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+18}{2*1}=\frac{8}{2}
=4
$